ESO 208A: Computational Methods in Engineering. Saumyen Guha

Transcription

1 ESO 208A: Computational Methods in Engineering Introduction, Error Analysis Saumyen Guha Department of Civil Engineering IIT Kanpur

2 What is Computational Methods or Numerical Methods in Engineering? Formulation of mathematical problems in such a way that numerical answers can be computed using arithmetic operations in a computer Computer can only perform: + - / All kinds of problems can be solved: linear and non-linear systems of equations; approximation and interpolation of data; differentiation and integration, ODE, PDE

3 What is Numerical Analysis? Mathematical method to analyze whether the solution obtained from a Numerical Method represents the solution of the original mathematical problem! A way to test for convergence! Convergence can be conditional or unconditional! A method cannot be used for a given problem without the analysis!

4 Example 1: Calculate the square root of 2. Mathematical Problem: x 2 = 2, x =? or Calculate the root of x 2 2 = 0 Square root of 2 (an irrational number) cannot be computed using simple arithmetic operations! Generate a Cauchy sequence that converges to the square root of 2. Some Numerical Methods: x n+1 = 2 x n x n+1 = 1 x 2 n + 2 x n x n+1 = x n+2 x n +1

5 Let s start all the methods with x 0 = 1 and compute for five iterations: x n+1 = 2 x n : x 1 = 2, x 2 = 1, x 3 = 2, x 4 = 1, x 5 = 2 x n+1 = 1 2 x n + 2 x n : x 1 = 1.5, x 2 = , x 3 = , x 4 = , x 5 = x n+1 = x n+2 x n +1 : x 1 = 1.5, x 2 = 1.4, x 3 = , x 4 = , x 5 = Could we have predicted this without having to perform computations?

6 Test for convergence of sequence (MTH 101): Sequence: x n+1 = 1 2 x n + 2 x n = 2x n x n+1 x n Since the root is real, x n Therefore, x n = 2x n x n+1 2 2x n x n+1 2, the sequence is bounded! for n 2, x n x n+1 = x n 1 2 x n x n 2 2 x n x n = 0, monotonically decreasing sequence! Monotonically decreasing sequence and bounded:- a Cauchy sequence that converges to the lower bound 2

7 Example 2: Solve the problem of the radioactive decay: dn dt = λn at t = 0, N = N 0 True Solution: N = N 0 e λt A Numerical Method: Euler Method (MTH 102) N i+1 = N i λhn i where, h = time step Δt Consider two different elements with λ = 0.1 and 2.2 Let us compare true analytical solutions with the numerical solutions for h = Δt = 1 for these two elements.

8 λ = 0.1 /day, N 0 = 10 6, h = Δt = 1, solution for 25 days There is some error but the numerical solution behaves like the true solution

9 λ = 2.2 /day, N 0 = 10 6, h = Δt = 1, solution for 25 days The error can grow to make the solution unacceptable! Numerical analysis could have told me this saving much time of programming and computation!

10 λ = 0.1 /day, N 0 = 10 6, varying h, solution for 25 days Numerical analysis could have told me this saving much time of programming and computation!

11 Take away message: Learning Numerical Analysis is as important as the Numerical Method for you to be able to apply the method for a given problem!! This is true irrespective of whether you are writing the program or using an existing program or package or software!

12 Errors and Error Analysis

13 Accuracy, Precision what are they? Accuracy: how closely a computed or measured value agrees with the True value opposite sense: inaccurate Precision: How closely individual computed or measured values agree with each other Opposite sense: imprecise

14 Define Error: True Value (a) = Approximate Value a + Error (ε) Absolute Error: ε = a a Relative Error: e = ε a = a a a Relative error is often expressed as (%) by multiplying (e) with 100. Absolute error can have sign as well as. If the error is computed with respect to the true value (if known), a prefix True is added. For an iterative process, the true value a is replaced with the previous iteration value and a prefix approximate is added. This is used for testing convergence of the iterative process.

15 Sources of Error in computation? Errors in the Input data: initial and boundary conditions, measured values of the parameters and constants in the model Round-off error: irrational numbers, product and division of two numbers, limited by the machine capability Truncation error: truncation of an infinite series, often arises in the design of the numerical method through approximation of the mathematical problem.

16 Error Analyses: Forward Error Analysis: Error in the Input Both computations are using the original Mathematical Model How an error in the given input propagates through the system model. (Robustness of the model) Backward Error Analysis: Error in the Output Computation using numerical algorithm Computation using mathematical problem Error in the Output Quantification of error resulting from interaction between round-off error and truncation error in the algorithm. (Robustness of the algorithm)

17 Forward Error Analysis: Single Variable Function: y = f(x). If an error is introduced in x, what is the error in y? x = x x y = y y = f x f x f x = f x + x = f x + xf x + x2 2! f x +.. Assuming the error to be small, the 2 nd and higher order terms are neglected. (a first order approximation!) y = f x f x xf x

18 Condition Number of the Problem (C p ): C p = Relative Error in y Relative Error in x = y y x x xf x f x = xf x x x f x Also: C p = As Δx 0, y y x = Τ x f x f x f x x = x Τ x f x f x+ x f x x C p = xf f x x C p < 1: problem is well-conditioned, error is attenuated C p > 1: problem is ill-conditioned, error is amplified C p = 1: neutral, error is translated

19 Examples of Forward Error Analysis and C p : Problem 1: y = e x ; y = xe x ; C p = x. The problem is well-conditioned for 0 x < 1; neutral at x = 1 and illconditioned for x > 1. Problem 2: Solve the following system of equations: x + αy = 1; αx + y = 0 Solving: x = 1 1 α 2 = x α ; x α = 2α 1 α 2 2 2α α 1 α C p = 2 2 2α 2 = 1 1 α 2 1 α 2 well-conditioned for α << 1 and ill-conditioned for α 1.

20 Forward Error Analysis: Function of Multiple Variables: y = f(x), where X is a vector, X = {x 0, x 1,.. x n }. If error is introduced in the x i s, what is the error in y? X = X X = f X = f X + x i n i=1 y = y y = f X f X x 0 x 0 x 1 x 1 = x n x n f ቤ x i An upper bound of y = σn i=1 X x 0 x 1 x n + HOT f σ n i=1 x i ቚ x i x i f x i ቚ X X

21 Bacward Error Analysis: Computation using numerical algorithm Computation using mathematical problem Error in the Output What is the perturbation required in the input in order to explain the error in the output if the computation is carried out by true mathematical function without any error. Hypothetical error in the Input computed using backward error analysis Error in the Output Both computations are using the original Mathematical Model

22 Floating point representation of a number = m b q m = mantissa; 1/b m 1 b = base; 2 for binary, 10 for decimal, 16 for hexadecimal q Q (set of rationals) If a machine (computer) rounds a number off to t decimal places: a = m 10 q ; a = m 10 q m m t Machine round-off unit(u) is the upper bound of the relative error in one round-off operation a a t 10 q a m 10 q t = u Condition Number of Algorithm (C a ): change necessary in the input data in order to explain the error in the final result expressed in terms of u.

23 Consider one floating point operation fl(x op y). op can be any of +, -,, /. By definition of u: fl x op y x op y x op y u Let s consider the op as : fl(x y) = x y(1 + δ) x y(1 + u) where δ u Example: A machine with t = 2, u = = 0.05 x = 0.30, y = 0.51, x y = 0.153; fl(x y) = 0.15 Relative error = ( )/0.153 = 0.02 u (= 0.05) For this operation: δ = -0.01; δ = 0.01 u (= 0.05)

24 Multiple Floating Point Operations: fl x 1 x 2 x 3 x n = x 1 x δ 1 x δ 2 x n 1 + δ n 1 δ i u for i = 1, 2, n 1 Relative error in the final computed value: e = fl x 1 x 2 x 3 x n x 1 x 2 x 3 x n x 1 x 2 x 3 x n = 1 + δ δ δ n u n 1 1 The quantity 1 + u n 1 1 may be approximated as 1.06(n-1)u for (n-1)u < 0.1

25 Example: Using a machine with 4-decimal place precision, t = 4, u = = Compute y = 1 + sin x 1 for x = 1 fl x = π = fl sin x = fl 1 + sin x = fl 1 + sin x = fl 1 + sin x 1 = True value of y using infinite precision is This is equivalent to performing the same computation in a machine with infinite precision with a starting x = Required relative change in x is ( )/ = Condition number of the algorithm is C a = / sin x If the algorithm is changed to y =, the same machine with 4-1+sin x+1 decimal place precision computes y = with C a = 0.4

26 Total Relative Error in the output: Recall: C p = Relative Error in the output Relative Error in the input Relative Error in the Output = C p Relative Error in the Input Relative Error in the Input has two components: User specified relative error in the input data = r (say) Error introduced due to truncation and round-off errors in the algorithm. Equivalent relative error in the input due to this is computed by backward error analysis as C a u Total relative error in the Input data = (r + C a u) Therefore, Relative Error in the Output = C p (r + C a u)